In a triangle of one vertex, the height of the bisector and the median are drawn. Distances from the other vertex

In a triangle of one vertex, the height of the bisector and the median are drawn. Distances from the other vertex to the base of the height of the bisector and the median are, respectively, 21cm 25cm 25.5cm. Calculate the perimeter of the triangle.

Consider a triangle ABC and let BH – height, BN – bisector,

BM is the median.

By the condition of the problem, it is known that:

AH = 21, AN = 25, AM = 25.5.

Note that since BM is the median, then AM = CM = 25.5.

Therefore, AC = AM + CM = 51.

Since N is the base of the bisector, then by the property of the bisector we have:

AB / AN = BC / CN,

AB / 25 = BC / 26, BC = 26/25 * AB.

By the cosine theorem, we have:

BC ^ 2 = AB ^ 2 + AC ^ 2 – 2 * AB * AC * cos (A).

Note that AB * cos (A) = AH = 21. Therefore:

(26/25) ^ 2 * AB ^ 2 = AB ^ 2 + 51 ^ 2 – 2 * 51 * 21,

51/25 ^ 2 * AB ^ 2 = 51 * 9,

AB ^ 2 = 25 ^ 2 * 9,

AB = 75.

BC = 26/25 * AB = 78.

Therefore, the perimeter

P = AB + BC + AC = 75 + 78 + 51 = 204.

Answer: 204 cm.